How to find the sum of the series $\sum_{k=2}^\infty \frac{1}{k^2-1}$? I have this problem :
$$S_n=\sum_{k=2}^\infty \frac{1}{k^2-1}$$
My solution
$$S_n=\sum_{k=2}^\infty \frac{1}{k^2-1} = -\frac{1}{2}\sum_{k=1}^\infty \frac{1}{k+1} -\frac{1}{k-1} = -\frac{1}{2}[(\frac{1}{3}-1)+(\frac{1}{4}-\frac{1}{2})+(\frac{1}{5}-\frac{1}{3})+(\frac{1}{6}-\frac{1}{4})+...]$$
I think that the sum should be $\frac{1}{2}$ since the limit of :
$$-\frac{1}{2}[-1+\frac{1}{k}+...+\frac{1}{n}] = \frac{1}{2}$$
But that wrong. Any ideas?
 A: Let $S_n=\displaystyle\sum_{k=1}^n\frac{1}{k^2-1}=\sum_{k=1}^n\frac{1}{2}\left[\frac{1}{k-1}-\frac{1}{k+1}\right]$
$\displaystyle\hspace{.2 in} =\frac{1}{2}\left[\bigg(\frac{1}{1}-\frac{1}{3}\bigg)+\bigg(\frac{1}{2}-\frac{1}{4}\bigg)+\bigg(\frac{1}{3}-\frac{1}{5}\bigg)+\cdots+\bigg(\frac{1}{n-2}-\frac{1}{n}\bigg)+\bigg(\frac{1}{n-1}-\frac{1}{n+1}\bigg)\right]$
$\hspace{.2 in}\displaystyle=\frac{1}{2}\left[1+\frac{1}{2}-\frac{1}{n}-\frac{1}{n+1}\right]$,
so $\displaystyle S=\lim_{n\to\infty}S_n=\lim_{n\to\infty}\frac{1}{2}\left[1+\frac{1}{2}-\frac{1}{n}-\frac{1}{n+1}\right]=\frac{3}{4}$.
A: your choice to use the "telescoping sum" technique is fine. what is wrong is just an arithmetic prob.
since the series is absolutely convergent try first rewriting as two sums:
$$
\sum_{k=1}^\infty \frac{1}{(2k)^2-1} + \sum_{k=1}^\infty \frac{1}{(2k+1)^2-1}
$$
and evaluate these separately. 
A: $$-\frac{1}{2}[(\frac{1}{3} - 1) + (\frac{1}{4} - \frac{1}{2}) + (\frac{1}{5} - \frac{1}{3}) + (\frac{1}{6} - \frac{1}{4})\ldots $$
$$-\frac{1}{2}[-1 - \frac{1}{2} + (\frac{1}{3} - \frac{1}{3}) + (\frac{1}{4} - \frac{1}{4}) + (\frac{1}{5} - \frac{1}{5}) \ldots $$
$$-\frac{1}{2}[-1 - \frac{1}{2}] = \frac{3}{4} $$
Notice that I am just rearranging to cancel all the positive terms as it is a telescoping sum. 
